Respuesta :
Answer is: 3,4
Chemical reaction: HNO₂ + NaOH → NaNO₂ + H₂O.
c₀(HNO₂) = 1,2 M = 1,2 mol/dm³.
c₀(NaNO₂) = 0,8 M = 0,8 mol/dm³.
V₀(HNO₂) = V₀(NaNO₂) = 1 dm³ = 1 L.
c₀(NaOH) = 0,5 M = 0,5 mol/dm³.
n₀(HNO₂)= 1,2 mol/dm³ · 1 dm³ = 1,2 mol.
n₀(NaNO₂) = 0,8 mol/dm³ · 1 dm³ = 0,8 mol.
V(NaOH) = 400 mL · 0,001 dm³/mL = 0,4 dm³.
n₀(NaOH) = c₀(NaOH) · V₀(NaOH).
n₀(NaOH) = 0,5 mol/dm³ · 0,4 dm³ = 0,2 mol.
n(HNO₂) = 1,2 mol - 0,2 mol = 1 mol.
n(NaNO₂) = 0,8 mol + 0,2 mol = 1 mol.
c(HNO₂) = 1 mol ÷ 1,4 dm³ = 0,714 mol/dm³.
c(NaNO₂) = 1 mol ÷ 1,4 dm³ = 0,714 mol/dm³.
pH = pKa + log (c(HNO₂) / c(NaNO₂)).
pH = 3,4 + log (0,714 mol/dm³ / 0,714 mol/dm³) = 3,4.
Chemical reaction: HNO₂ + NaOH → NaNO₂ + H₂O.
c₀(HNO₂) = 1,2 M = 1,2 mol/dm³.
c₀(NaNO₂) = 0,8 M = 0,8 mol/dm³.
V₀(HNO₂) = V₀(NaNO₂) = 1 dm³ = 1 L.
c₀(NaOH) = 0,5 M = 0,5 mol/dm³.
n₀(HNO₂)= 1,2 mol/dm³ · 1 dm³ = 1,2 mol.
n₀(NaNO₂) = 0,8 mol/dm³ · 1 dm³ = 0,8 mol.
V(NaOH) = 400 mL · 0,001 dm³/mL = 0,4 dm³.
n₀(NaOH) = c₀(NaOH) · V₀(NaOH).
n₀(NaOH) = 0,5 mol/dm³ · 0,4 dm³ = 0,2 mol.
n(HNO₂) = 1,2 mol - 0,2 mol = 1 mol.
n(NaNO₂) = 0,8 mol + 0,2 mol = 1 mol.
c(HNO₂) = 1 mol ÷ 1,4 dm³ = 0,714 mol/dm³.
c(NaNO₂) = 1 mol ÷ 1,4 dm³ = 0,714 mol/dm³.
pH = pKa + log (c(HNO₂) / c(NaNO₂)).
pH = 3,4 + log (0,714 mol/dm³ / 0,714 mol/dm³) = 3,4.
The new pH of given solutionis [tex]\boxed{3.53}[/tex].
Further Explanation:
The aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid is termed as buffer solution. These solutions offer strong resistance to any change in their pH on addition of small quantity of strong acid or base.
Henderson-Hasselbalch equation:
This equation helps in determining the pH of buffer solution. Its mathematical form is given as follows:
[tex]{\text{pH}} = {\text{p}}{K_{\text{a}}} + {\text{log}}\dfrac{{\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}}[/tex] …… (1)
Here,
[tex]\left[ {{{\text{A}}^ - }} \right][/tex] is concentration of conjugate base.
[HA] is concentration of acid.
Given mixture is a buffer solution of [tex]{\text{HN}}{{\text{O}}_{\text{2}}}[/tex] and [tex]{\text{NaN}}{{\text{O}}_{\text{2}}}[/tex]. Therefore Henderson-Hasselbalch equation becomes as follows:
[tex]{\text{pH}} = {\text{p}}{K_{\text{a}}} + {\text{log}}\dfrac{{\left[ {{\text{NaN}}{{\text{O}}_{\text{2}}}} \right]}}{{\left[ {{\text{HN}}{{\text{O}}_{\text{2}}}} \right]}}[/tex] …… (2)
Initial moles of [tex]{\text{HN}}{{\text{O}}_{\text{2}}}[/tex] can be calculated as follows:
[tex]\begin{aligned}{\text{Moles of HN}}{{\text{O}}_2} &= \left( {1.2{\text{ M}}} \right)\left( {{\text{1 L}}} \right)\\&= 1.2{\text{ mol}} \\\end{aligned}[/tex]
Initial moles of [tex]{\text{NaN}}{{\text{O}}_{\text{2}}}[/tex] can be calculated as follows:
[tex]\begin{aligned}{\text{Moles of NaN}}{{\text{O}}_{\text{2}}} &= \left( {0.8{\text{ M}}} \right)\left( {{\text{1 L}}} \right)\\&= 0.8{\text{ mol}} \\\end{aligned}[/tex]
Moles of NaOH can be calculated as follows:
[tex]\begin{aligned}{\text{Moles of NaOH}} &= \left( {0.5{\text{ M}}} \right)\left( {{\text{400 mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right) \\&= 0.2{\text{ mol}} \\\end{aligned}[/tex]
When addition of 0.2 moles of NaOH is done to the buffer solution, 0.2 moles of [tex]{\text{HN}}{{\text{O}}_{\text{2}}}[/tex] is neutralized while the same amount of [tex]{\text{NaN}}{{\text{O}}_{\text{2}}}[/tex] is formed. Since volumes are additive, total volume can be calculated as follows:
[tex]\begin{aligned}{\text{Total volume of solution}} &= \left( {1 + \left( {400{\text{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right)} \right){\text{ L}} \\ &= {\text{1}}{\text{.4 L}} \\\end{aligned}[/tex]
Therefore concentration of [tex]{\text{HN}}{{\text{O}}_{\text{2}}}[/tex] can be calculated as follows:
[tex]\begin{aligned}\left[ {{\text{HN}}{{\text{O}}_{\text{2}}}} \right] &= \frac{{\left( {1.2 - 0.2} \right){\text{ mol}}}}{{1.4{\text{ L}}}}\\&= 0.714{\text{ M}} \\\end{aligned}[/tex]
Therefore concentration of [tex]{\text{NaN}}{{\text{O}}_{\text{2}}}[/tex] can be calculated as follows:
[tex]\begin{aligned}\left[ {{\text{NaN}}{{\text{O}}_{\text{2}}}} \right] &= \frac{{\left( {1.2 + 0.2} \right){\text{ mol}}}}{{1.4{\text{ L}}}} \\ &= 1{\text{ M}} \\\end{aligned}[/tex]
Substitute 0.714 M for [tex]\left[ {{\text{HN}}{{\text{O}}_{\text{2}}}} \right][/tex], 1 M for [tex]\left[ {{\text{NaN}}{{\text{O}}_{\text{2}}}} \right][/tex] and 3.4 for [tex]{\text{p}}{K_{\text{a}}}[/tex] in equation (2).
[tex]\begin{aligned} {\text{pH}} &= 3.4 + {\text{log}}\left( {\frac{{{\text{1 M}}}}{{0.714{\text{ M}}}}} \right) \\&= 3.54 \\\end{aligned}[/tex]
Learn more:
- The reason for the acidity of water https://brainly.com/question/1550328
- Reason for the acidic and basic nature of amino acid. https://brainly.com/question/5050077
Answer details:
Grade: High School
Subject: Chemistry
Chapter: Acid, base and salts
Keywords: pH, buffer, pKa, NaNO2, HNO2, 3.4, 1 M, 0.714 M, concentration, total volume of solution, 1.2 mol, 0.8 mol, 0.2 mol, 3.54.
